MHT CET · Physics · Dual Nature of Matter
The maximum velocity of the photoelectron emitted by the metal surface is 'V'. Charge and mass of the photoelectron is denoted by 'e' and 'm' respectivley. The stopping potential in volt is
- A \(\frac{V^{2}}{\left(\frac{m}{e}\right)}\)
- B \(\frac{\mathrm{V}^{2}}{2\left(\frac{\mathrm{e}}{\mathrm{m}}\right)}\)
- C \(\frac{V^{2}}{\left(\frac{e}{m}\right)}\)
- D \(\frac{\mathrm{V}^{2}}{2\left(\frac{\mathrm{m}}{\mathrm{e}}\right)}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{V}^{2}}{2\left(\frac{\mathrm{e}}{\mathrm{m}}\right)}\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{2} \mathrm{mv}^{2}=\mathrm{eV}_{\mathrm{s}}\)
\(\therefore \mathrm{V}_{\mathrm{s}}=\frac{1}{2}\left(\frac{\mathrm{m}}{\mathrm{e}}\right) \mathrm{v}^{2}=\frac{\mathrm{v}^{2}}{2\left(\frac{\mathrm{e}}{\mathrm{m}}\right)}\)
\(\therefore \mathrm{V}_{\mathrm{s}}=\frac{1}{2}\left(\frac{\mathrm{m}}{\mathrm{e}}\right) \mathrm{v}^{2}=\frac{\mathrm{v}^{2}}{2\left(\frac{\mathrm{e}}{\mathrm{m}}\right)}\)
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